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Obviously no. If you manage to compute an eigenvector $\mathbf{v}$ of a matrix $A$ then to automatically get the associated eigenvalue $\lambda$ by the formula $A\mathbf{v}=\lambda\mathbf{v}$.
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Fernando MuroSep 23 '12 at 9:53

@Fernando: the comment is a bit tautological, since this is exactly what the OP is proposing to do.
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Igor RivinSep 23 '12 at 18:45

If we know that the rows all have the same sum (but not what that sum is) then we would essentially find it by multiplying by the corresponding eigenvector, $\mathbb{j}$, the all $1$'s vector. This will be the largest eigenvalue provided that the entries are non-negative.

One way this could happen (but not the only one) is if the rows are identical or merely each is a permutation of the first.

If certain rows are equal then we know that $0$ is a eigenvalue although we never "computed" it. Then we do know an eigenvector.

In a circulant matrix we know all the eigenvectors (not just $\mathbb{j}$) and we essentially use them to compute the corresponding eigenvalues.